MHB Perpendicular vectors, triangle, tetrahedron

[furor celtica]

Prove that, if (c - b).a = 0 and (c - a).b = 0, then (b - a).c = 0. Show that this can be used to prove the following geometric results:
a. The lines through the vertices of a triangle ABC perpendicular to the opposite sides meet in a point.
b. If the tetrahedron OABC has two pairs of perpendicular opposite edges, the third pair of edges is perpendicular.
Prove that also, in both cases, (OA)^2 + (BC)^2 = (OB)^2 + (CA)^2 = (OC)^2 + (AB)^2So for the very first task I proceeded by attributing coordinates to each vector: a = (x(1), y(1), z(1)), b = (x(2), y(2), z(2), c = (x(3), y(3), z(3)), where the numbers are actually at the bottom left of the coordinates, but I don't know how to use that notation here.
Anyway I could post all my work but it would take a long time; I didn't have much of a problem at all with the first task and proved that (b - a).c = 0. It took a while though, as you can imagine; is there another way to solve problems like this, or does one always have to use attributed coordinates and deal with those?

With a. and b. I'm stuck, however.
a. I usually would try to solve a question like this by taking into account the end result and what exactly I'm supposed to end up with (and often work backwards to return to the given results), but here I'm not sure at all what you end up with algebraically when three vectors meet. What am I looking for? Also I'm confused on how to find the vectors that are perpendicular to the opposite sides, how to formulate them with regard to the coordinates of A, B and C.
b. Here I immediately became confused as to what is meant by 'opposite edges'. But besides that it seems to be a fairly straightforward variation of the very first task, am I correct?
The last task seems simple as well, as I would just take the attributed coordinates and reformulate the magnitudes of the given vectors to be equivalent, perhaps with some help from the result in the first task. I haven't tackled this one yet.

 

[earboth]

Prove that, if (c - b).a = 0 and (c - a).b = 0, then (b - a).c = 0. Show that this can be used to prove the following geometric results:
a. The lines through the vertices of a triangle ABC perpendicular to the opposite sides meet in a point.
b. If the tetrahedron OABC has two pairs of perpendicular opposite edges, the third pair of edges is perpendicular.
Prove that also, in both cases, (OA)^2 + (BC)^2 = (OB)^2 + (CA)^2 = (OC)^2 + (AB)^2So for the very first task I proceeded by attributing coordinates to each vector: a = (x(1), y(1), z(1)), b = (x(2), y(2), z(2), c = (x(3), y(3), z(3)), where the numbers are actually at the bottom left of the coordinates, but I don't know how to use that notation here.
Anyway I could post all my work but it would take a long time; I didn't have much of a problem at all with the first task and proved that (b - a).c = 0. It took a while though, as you can imagine; is there another way to solve problems like this, or does one always have to use attributed coordinates and deal with those?

With a. and b. I'm stuck, however.
a. I usually would try to solve a question like this by taking into account the end result and what exactly I'm supposed to end up with (and often work backwards to return to the given results), but here I'm not sure at all what you end up with algebraically when three vectors meet. What am I looking for? Also I'm confused on how to find the vectors that are perpendicular to the opposite sides, how to formulate them with regard to the coordinates of A, B and C.
b. Here I immediately became confused as to what is meant by 'opposite edges'. But besides that it seems to be a fairly straightforward variation of the very first task, am I correct?
The last task seems simple as well, as I would just take the attributed coordinates and reformulate the magnitudes of the given vectors to be equivalent, perhaps with some help from the result in the first task. I haven't tackled this one yet.

Only to b)

An edge of a tetrahedron is the intersection of two triangular sides. The edge which doesn't belog to these two triangles is opposite of the intersection of the triangles.

I've attached a sketch. Edge and opposite edge are drawn in the same colour. To get an idea how such a tetrahedron could look like take a cube and cut off one vertex.

EDIT: I've attached a more detailed sketch. Maybe this helps better!

 

[furor celtica]

thanks, but I'm firstly stuck on a.!

 

[earboth]

Prove that, if (c - b).a = 0 and (c - a).b = 0, then (b - a).c = 0. Show that this can be used to prove the following geometric results:
a. The lines through the vertices of a triangle ABC perpendicular to the opposite sides meet in a point.
...
With a. and b. I'm stuck, however.
a. I usually would try to solve a question like this by taking into account the end result and what exactly I'm supposed to end up with (and often work backwards to return to the given results), <-- that's exactly the best way to do this question
but here I'm not sure at all what you end up with algebraically when three vectors meet. What am I looking for? Also I'm confused on how to find the vectors that are perpendicular to the opposite sides, how to formulate them with regard to the coordinates of A, B and C.
...

1. Two sides of a triangle are not parallel. The heights of these two sides of the triangle must intersect in a point H. You now have to show that the line from the vertex (that's the point of intersection of the two known sides) through H is perpendicular to the 3rd side of the triangle.

2. I'm referring to the attached sketch.

$ \overrightarrow{HA} \cdot \overrightarrow{BC} = 0~\wedge~\overrightarrow{HC} \cdot \overrightarrow{AB} = 0 $
and:
$ \overrightarrow{BC} = \overrightarrow{HC} - \overrightarrow{HB} ~\wedge~\overrightarrow{AB} = \overrightarrow{HB} - \overrightarrow{HA} $

3. Plug in:

$ \overrightarrow{HA} \cdot (\overrightarrow{HC} - \overrightarrow{HB}) = 0 $ and
$ \overrightarrow{HC} \cdot (\overrightarrow{HB} - \overrightarrow{HA}) = 0 $

Expand the brackets and add the LHS of the equations:

4.
$ \overrightarrow{HC} \cdot \overrightarrow{HB} - \overrightarrow{HA} \cdot \overrightarrow{HB} = 0 $

$ (\overrightarrow{HC} - \overrightarrow{HA}) \cdot \overrightarrow{HB} = 0 $

$ \overrightarrow{AC} \cdot \overrightarrow{HB} = 0 $

q.e.d